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51 won't contribute to the number of zeros at the end of the number, therefore all zeros will come from 49!.

Trailing zeros in 49!: \(\frac{49}{5}+\frac{49}{5^2}=9+1=10\)

Answer: B.

THEORY:

Trailing zeros:

Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.

For example 125,000 has 3 trailing zeros;

The number of trailing zeros n!, the factorial of a non-negative integer \(n\), can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where \(k\) must be chosen such that \(5^{(k+1)} \gt n\)

It's easier if we consider an example:

How many zeros are in the end (after which no other digits follow) of 32!?

\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\). Notice that the last denominator (\(5^2\)) must be less than 32. Also notice that we take into account only the quotient of the division, that is \(\frac{32}{5}=6\).

So there are 7 zeros in the end of 32!.

Another example, how many trailing zeros does 125! have?

\(\frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31\),

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

Bunuel why 51 won't contribute to the number of zeros at the end of the number ? i dont get

51 won't contribute to the number of zeros at the end of the number, therefore all zeros will come from 49!.

Trailing zeros in 49!: \(\frac{49}{5}+\frac{49}{5^2}=9+1=10\)

Answer: B.

THEORY:

Trailing zeros:

Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.

For example 125,000 has 3 trailing zeros;

The number of trailing zeros n!, the factorial of a non-negative integer \(n\), can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where \(k\) must be chosen such that \(5^{(k+1)} \gt n\)

It's easier if we consider an example:

How many zeros are in the end (after which no other digits follow) of 32!?

\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\). Notice that the last denominator (\(5^2\)) must be less than 32. Also notice that we take into account only the quotient of the division, that is \(\frac{32}{5}=6\).

So there are 7 zeros in the end of 32!.

Another example, how many trailing zeros does 125! have?

\(\frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31\),

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

Bunuel why 51 won't contribute to the number of zeros at the end of the number ? i dont get

\(\frac{49}{5} = 9\)

\(\frac{51}{5^2}\) = 2

\(9+2 =11\)

The formula works only when we have a factorial !when we have an integer , it has to be factorised.

consider 6!6!= 6*5*4*3*2*1 no of zeroes = no of 5*2 pairs we have only one in this case , so no of trailing zeroes = 1using the formula 6/5=1

now consider 66=2*3doesn't have a five in it , and therefore doesn't have a trailing zero

similarly 51 is an integer factors of 51 are 1,3,17,51no 5's or 2's , so no trailing zeroes